Operating system concept problem

With a 2^32 address space and 4K ( 2^12 ) page sizes, this leave 2^20 entries in the page table. At 4 bytes per entry, this amounts to a 4 MB page table, which is too large to reasonably keep in contiguous memory. ( And to swap in and out of memory with each process switch. ) Note that with 4K pages, this would take 1024 pages just to hold the page table!

Please explain me how they are calculating 1024 pages at the end? what is 4bytes per entry in this? Thanks in advance.

That is: you have 4GB of total space memory. (The question is wrong because it says 4MB and must say 4GB).

2^32/1024/4 = 2^32/4096 = 1048576 pages of memory.

2^20 = 1048576. That is, you have 2^20 pages of memory.

2^20 = 1048576 = 1024x1024. That is you have 1024 k pages of memory.

You have 1M pages of memory and each page has 4k bytes.

If one entry uses 4 bytes, you have 4096/4 = 1024 entries per memory page.
So, you have 1k entries per page.

With all of this numbers, I think the problem is wrong formulated. It must say something like:

With a 2^32 address space and 4K (2^12) page sizes, this leave 2^20 pages in the table. At 4 bytes per entry, this amounts to a 4 GB page table, which is too large to reasonably keep in contiguous memory. (And to swap in and out of memory with each process switch.) Note that with 4K page sizes, this would take 1024 k pages (=1M pages) just to hold the total table!

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